# Spivak's proof of Inverse Function Theorem

I am having trouble with Spivak's proof of the Inverse Function Theorem in his Calculus on Manifolds:

2-11 Theorem (Inverse Function Theorem). Suppose that $f: \mathbb{R}^n\to\mathbb{R}^n$ is continuously differentiable in an open set containing $a$, and det $f'(a)\neq 0$. Then there is an open set $V$ containing $a$ and an open set $W$ containing $f(a)$ such that $f:V\to W$ has a continuous inverse $f^{-1}:W\to V$ which is differentiable and for all $y\in W$ satisfies $$(f^{-1})'(y) = [f'(f^{-1}(y))]^{-1}$$ Proof. Let $\lambda$ be the linear transformation $Df(a)$. Then $\lambda$ is non-singular, since det $f'(a)\neq 0$. Now $D(\lambda^{-1}\circ f)(a) = D(\lambda^{-1})(f(a))\circ Df(a) = \lambda^{-1}\circ Df(a)$ is the identity linear transformation. If the theorem is true for $\lambda^{-1}\circ f$, it is clearly true for $f$...

How is the theorem true for $f$ if it is true for $\lambda^{-1}\circ f$?

• I have not read Spivak's book, but I would have guessed $\lambda$ is the constant map $x\rightarrow \lambda x$, which does not influence $f$'s differentiablity since $\lambda f$ is just multiplying $f$ by a constant. – Bombyx mori Sep 1 '12 at 20:35

$$\lambda\colon \mathbb{R}^n\to\mathbb{R}^n$$ is a bijection and $$\lambda$$ and $$\lambda^{-1}$$ are both continuously differentiable. Note that $$\lambda'(z) = \lambda$$ for all $$z \in \mathbb{R}$$.

Let $$g = \lambda^{-1}\circ f$$. Suppose the theorem is true for $$g$$. Then there is an open set $$V'$$ containing $$a$$ and an open set $$W'$$ containing $$g(a)$$ such that $$g:V'\to W'$$ has a continuous inverse $$g^{-1}:W'\to V'$$ which is differentiable and for all $$y\in W'$$ satisfies $$(g^{-1})'(y) = [g'(g^{-1}(y))]^{-1}$$ Then $$\lambda(W')$$ is open and $$f = \lambda\circ g:V'\to \lambda(W')$$ has a continuous inverse $$g^{-1}\circ \lambda^{-1}:\lambda(W')\to V'$$.

By the chain rule, for all $$z \in \lambda(W')$$, $$(f^{-1})'(z) = (g^{-1}\circ\lambda^{-1})'(z) = (g^{-1})'(\lambda^{-1}(z))\circ \lambda^{-1} = [g'(g^{-1}(\lambda^{-1}(z)))]^{-1}\circ \lambda^{-1} = [g'(f^{-1}(z))]^{-1}\circ \lambda^{-1} = [\lambda\circ g'(f^{-1}(z)]^{-1} = [f'(f^{-1}(z))]^{-1}$$

• Why is $\lambda$ a bijection? – mathemather May 27 '18 at 12:07
• $\lambda$ is non-singular (since $\det f'(a) \not= 0$). Hence it estabilish a one-to-one correspondence between the function domain and image (i.e. it is a bijection) – Antonio Horta Ribeiro Mar 28 at 7:53

Suppose the theorem holds for all functions $f$ such that $Df(a) = I$. What Spivak has shown is that for $\lambda = Df(a)$, $D(\lambda^{-1} \circ f(a)) = I$. By the assumption above, the theorem must be true for $\lambda^{-1}\circ f$. That means there exists $g$ which is the continuously differentiable inverse of $\lambda^{-1} \circ f$. This implies that $(g \circ \lambda^{-1}) \circ f = I$. Now $g \circ \lambda^{-1}$ is continuously differentiable, and so the theorem is true for $f$.

• I dont understand the last statement. How does $g \circ \lambda ^{-1}$being continuously differentiable imply the theorem is true for $f$? – mathemather May 27 '18 at 12:11
• Notice how he placed the parentheses around $g\circ\lambda^{-1}$. This has no effect on computing the compositions; he simply did it to show you that $g\circ\lambda^{-1}$ is in fact the inverse of $f$ that we are looking for! And in addition, this inverse is continuously differentiable since it is just the map $g$ and we generated $g$ as a result of applying the Inverse function Theorem to $\lambda^{-1} \circ f(a)$, so $g$ certainly satisfies all the properties that we'd want from our inverse. – john fowles Jun 7 '18 at 0:01